| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 18205 | Accepted: 3960 |
Description
The farm has F (1 <= F <= 200) fields on which the cows graze.
A set of P (1 <= P <= 1500) paths connects them. The paths are
wide, so that any number of cows can traverse a path in either
direction.
Some of the farm's fields have rain shelters under which the cows
can shield themselves. These shelters are of limited size, so a single
shelter might not be able to hold all the cows. Fields are small
compared to the paths and require no time for cows to traverse.
Compute the minimum amount of time before rain starts that the siren must be sounded so that every cow can get to some shelter.
Input
* Lines 2..F+1: Two space-separated integers that describe a field.
The first integer (range: 0..1000) is the number of cows in that field.
The second integer (range: 0..1000) is the number of cows the shelter
in that field can hold. Line i+1 describes field i.
* Lines F+2..F+P+1: Three space-separated integers that describe a
path. The first and second integers (both range 1..F) tell the fields
connected by the path. The third integer (range: 1..1,000,000,000) is
how long any cow takes to traverse it.
Output
Line 1: The minimum amount of time required for all cows to get under a
shelter, presuming they plan their routes optimally. If it not possible
for the all the cows to get under a shelter, output "-1".
Sample Input
3 4
7 2
0 4
2 6
1 2 40
3 2 70
2 3 90
1 3 120
Sample Output
110
Hint
In 110 time units, two cows from field 1 can get under the shelter
in that field, four cows from field 1 can get under the shelter in field
2, and one cow can get to field 3 and join the cows from that field
under the shelter in field 3. Although there are other plans that will
get all the cows under a shelter, none will do it in fewer than 110 time
units.
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <time.h>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define inf 0x3f3f3f3f
#define mod 10000
typedef long long ll;
using namespace std;
const int N=;
const int M=;
int power(int a,int b,int c){int ans=;while(b){if(b%==){ans=(ans*a)%c;b--;}b/=;a=a*a%c;}return ans;}
struct man
{
int c,f;
}w[N][N];
int dis[N],n,m;
int t,cnt,maxn=,ans;
ll cost[N][N];
int c[N],f[N];
bool bfs()
{
queue<int>q;
memset(dis,,sizeof(dis));
q.push();
dis[]=;
while(!q.empty() && !dis[t]){
int v=q.front();q.pop();
for(int i=;i<=t;i++){
//if(i==t)printf("w[i][t].c=%d\n",w[i][t].c);
if(!dis[i]&&w[v][i].c>w[v][i].f){
q.push(i);
dis[i]=dis[v]+;
}
}
}
return dis[t]!=;
}
int dfs(int cur,int cp)
{
if(cur==t||cp==)return cp;
int tmp=cp,tt;
for(int i=;i<=t;i++){
if(dis[i]==dis[cur]+ &&w[cur][i].c>w[cur][i].f){
tt=dfs(i,min(w[cur][i].c-w[cur][i].f,tmp));
w[cur][i].f+=tt;
w[i][cur].f-=tt;
tmp-=tt;
}
}
return cp-tmp;
}
void dinic()
{
ans=;
while(bfs())ans+=dfs(,inf);
}
void Floyd()
{
for(int k=;k<=n;k++){
for(int i=;i<=n;i++){
for(int j=;j<=n;j++){
if(i!=j)cost[i][j]=min(cost[i][j],cost[i][k]+cost[k][j]);
else cost[i][j]=;
}
}
}
}
void Build(ll x)
{
memset(w,,sizeof(w));
for(int i=;i<=n;i++)w[][i].c=c[i];
for(int i=n+;i<=*n;i++)w[i][t].c=f[i-n];
for(int i=;i<=n;i++)for(int j=n+;j<t;j++)if(cost[i][j-n]<=x)w[i][j].c=inf;
}
int main(){
cin>>n>>m;
memset(cost,inf,sizeof(cost));
ll l=,r=;
t=n*+;
for(int i=;i<=n;i++){cin>>c[i]>>f[i];maxn+=c[i];}
int a,b;ll val;
while(m--){
cin>>a>>b>>val;
r+=val;
cost[a][b]=cost[b][a]=min(cost[a][b],val);
}
Floyd();
bool flag=false;
while(l<r){
ll mid=(l+r)/;
Build(mid);
dinic();
if(ans>=maxn)r=mid,flag=true;
else l=mid+;
}
if(flag) cout<<r<<endl;
else puts("-1");
return ;
}
AC代码
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <time.h>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define inf 0x3f3f3f3f
#define mod 10000
typedef long long ll;
using namespace std;
const int N=;
const int M=;
int power(int a,int b,int c){int ans=;while(b){if(b%==){ans=(ans*a)%c;b--;}b/=;a=a*a%c;}return ans;}
struct man
{
int c,f;
}w[N][N];
int dis[N],n,m;
int t,cnt,maxn=,ans;
ll cost[N][N];
int c[N],f[N];
bool bfs()
{
queue<int>q;
memset(dis,,sizeof(dis));
q.push();
dis[]=;
while(!q.empty() && !dis[t]){
int v=q.front();q.pop();
for(int i=;i<=t;i++){
//if(i==t)printf("w[i][t].c=%d\n",w[i][t].c);
if(!dis[i]&&w[v][i].c>w[v][i].f){
q.push(i);
dis[i]=dis[v]+;
}
}
}
return dis[t]!=;
}
int dfs(int cur,int cp)
{
if(cur==t||cp==)return cp;
int tmp=cp,tt;
for(int i=;i<=t;i++){
if(dis[i]==dis[cur]+ &&w[cur][i].c>w[cur][i].f){
tt=dfs(i,min(w[cur][i].c-w[cur][i].f,tmp));
w[cur][i].f+=tt;
w[i][cur].f-=tt;
tmp-=tt;
}
}
return cp-tmp;
}
void dinic()
{
ans=;
while(bfs())ans+=dfs(,inf);
}
void Floyd()
{
for(int k=;k<=n;k++){
for(int i=;i<=n;i++){
if(cost[i][k]!=inf){
for(int j=;j<=n;j++){
cost[i][j]=min(cost[i][j],cost[i][k]+cost[k][j]);
}
}
}
}
}
void Build(ll x)
{
memset(w,,sizeof(w));
for(int i=;i<=n;i++)w[][i].c=c[i];
for(int i=n+;i<=*n;i++)w[i][t].c=f[i-n];
for(int i=;i<=n;i++)for(int j=n+;j<t;j++)if(cost[i][j-n]<=x)w[i][j].c=inf;
}
int main(){
cin>>n>>m;
for(int i=;i<=n;i++){
for(int j=;j<=n;j++){
if(i==j)cost[i][j]=;
else cost[i][j]=inf;
}
}
ll l=,r=;
t=n*+;
for(int i=;i<=n;i++){cin>>c[i]>>f[i];maxn+=c[i];}
int a,b;ll val;
while(m--){
cin>>a>>b>>val;
r+=val;
cost[a][b]=cost[b][a]=min(cost[a][b],val);
}
Floyd();
bool flag=false;
while(l<r){
ll mid=(l+r)/;
Build(mid);
dinic();
if(ans>=maxn)r=mid,flag=true;
else l=mid+;
}
if(flag) cout<<r<<endl;
else puts("-1");
return ;
}
上面的是WA代码，不知道为什么错了。